Square root spiral

The governing idea of a square root spiral is the repeated use of right triangles and the Pythagorean theorem. Starting with a suitable unit length, a perpendicular segment of length 1 is added to the existing hypotenuse. If the old hypotenuse is √n, then the new one satisfies h² = (√n)² + 1² = n + 1, so h = √(n + 1). Thus the hypotenuses represent successive square roots such as √2, √3, √4, and so on. Option A is correct. Option B is false because the hypotenuse changes; option C is false because the triangles are right-angled, not equilateral; and option D is geometrically impossible for these constructions.

In a square root spiral, each new right triangle uses the previous hypotenuse and a perpendicular side of 1 unit. After \(\sqrt{9}\), the hypotenuse is \(\sqrt{9+1}=\sqrt{10}\). Using \(\sqrt{9}\) as the new side is incorrect. Exam tip: the added side is always 1 unit.

In a square root spiral, each new right triangle has one side of length 1, so its hypotenuse becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Hence \(\sqrt{11}\) can be represented. \(\pi\) is not a successive hypotenuse in this construction. Exam tip: look for square roots of non-square natural numbers.

Since 17 is not a perfect square, \(\sqrt{17}\) cannot be expressed as a ratio of two integers. Therefore, this length on the spiral represents an irrational number. Exam tip: first check whether the radicand is a perfect square.

The new side of length 1 is drawn perpendicular to the previous hypotenuse, so the two lengths are the legs of a right triangle. The previous hypotenuse has length \sqrt{n}. Applying the Pythagorean theorem gives the square of the new hypotenuse as (\sqrt{n})^2+1^2=n+1.

Taking the positive square root, because a length is positive, gives the new hypotenuse \sqrt{n+1}. This is why option A is correct. Option B incorrectly treats the new length as \sqrt{n}+1; in general, adding lengths is not the same as adding their squares under a perpendicular construction. The result depends specifically on the right angle and the Pythagorean theorem.

At each step, a new side of length 1 is drawn perpendicular to the previous hypotenuse. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). The triangles are not equilateral. Exam tip: identify the old hypotenuse and unit side.

The governing construction rule is that a unit perpendicular is added to the current hypotenuse. If the current hypotenuse is √n, the next hypotenuse has length √(n + 1), because h² = (√n)² + 1² = n + 1. To obtain √32, the preceding value must therefore have n + 1 = 32. Solving gives n = 31, so the immediately previous hypotenuse is √31. Option B is correct. √30 would lead to √31, not √32. √32 is the target itself rather than the previous hypotenuse, while √33 comes after the target in the sequence. The answer follows from the construction, not from subtracting ordinary lengths.

For the new right triangle, hypotenuse² = \((\sqrt{7})^2 + 1^2 = 7+1=8\), so the hypotenuse is \(\sqrt{8}\). It is not \(2\sqrt{7}\), as lengths are not simply doubled. Exam tip: apply Pythagoras’ theorem at every step.

In a square root spiral, the hypotenuse of one triangle becomes a side of the next, with a perpendicular side of length 1 added. By Pythagoras, the new hypotenuse squared is the previous hypotenuse squared plus 1. Exam tip: identify the new unit perpendicular side.

Since 10 is not a perfect square, \(\sqrt{10}\) cannot be expressed as a ratio of two integers, so it is irrational. Its spiral length is about 3.16. Exam tip: square roots of perfect squares are rational.

In a square root spiral, each new right triangle is constructed by taking the hypotenuse of the previous triangle as one of its sides. The direction of the new side keeps changing, so the entire construction appears to turn like a spiral. Option B is incorrect because the line segments are not parallel. Exam tip: identify the successive right triangles and their changing direction in a square root spiral.

A square root spiral advances by one under the radical at each step. More precisely, if a right triangle has an existing hypotenuse √n and a new perpendicular side of length 1, the Pythagorean theorem gives the next hypotenuse as √(n + 1). For the target √50, the immediately preceding value must satisfy n + 1 = 50, so n = 49. Therefore the hypotenuse formed before √50 is √49, making option B correct. √48 is two steps earlier, √50 is the requested target rather than its predecessor, and √51 is one step later. Although √49 simplifies to 7, it is written as √49 here to show its position in the spiral sequence.

The governing rule is obtained from the Pythagorean theorem. A unit segment is drawn perpendicular to the existing hypotenuse. If that hypotenuse is √11, then the new hypotenuse h satisfies h² = (√11)² + 1² = 11 + 1 = 12. Taking the positive length gives h = √12. Therefore option C is correct. √10 represents the preceding member of the sequence, and √11 is the original hypotenuse rather than the new one. √22 would result from a different calculation and has no basis in adding a perpendicular unit segment. The spiral therefore moves from √11 to √12, just as it moves from √n to √(n + 1) at every stage.

For the new right triangle, hypotenuse² = \((\sqrt{5})^2+1^2=5+1=6\). Hence the new hypotenuse is \(\sqrt{6}\) units. \(\sqrt{10}\) would result if the other side were also \(\sqrt{5}\). Exam tip: square the radical before applying Pythagoras’ theorem.

In a square root spiral, a perpendicular side of length 1 is drawn to the previous hypotenuse. By Pythagoras’ theorem, the new hypotenuse becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. It is not an equilateral triangle. Exam tip: remember that the added perpendicular side is always 1 unit.

For positive numbers, the square-root function is increasing. Since \(11>10\), we get \(\sqrt{11}>\sqrt{10}\), so Q is farther from the origin. Exam tip: compare the numbers inside the roots first.

The important concept is that the new hypotenuse is determined by the sum of squares, not by ordinary addition of the two side lengths. Starting with √2 and adding a perpendicular side of length 1, the Pythagorean theorem gives h² = (√2)² + 1² = 2 + 1 = 3, so h = √3. It is not correct to write √2 + 1 = √3; numerically, √2 + 1 is about 2.414, whereas √3 is about 1.732. Thus option A identifies the misconception to avoid. Options B, C, and D describe the actual construction and reasoning used to obtain √3, so they are not wrong ideas.

The direct answer is A: set square. A square root spiral is made by drawing repeated right triangles, usually adding a perpendicular side of length 1 each time. A set square has a fixed right-angle edge, so it helps the student draw a precise 90-degree angle. Step by step: place the ruler along the existing line; place the set square against it; draw the perpendicular; measure the required unit length; join the new endpoint to form the hypotenuse. Option A is correct. Option B, a balance, measures mass. Option C, a clock, measures or shows time. Option D, a measuring cylinder, measures liquid volume. None of these normally constructs a right angle. Exam cue: set square means accurate 90-degree construction.

In a square root spiral, each new hypotenuse is formed using the previous hypotenuse and a perpendicular unit side. Thus, \(\sqrt{5}\) comes from legs \(2\) and \(1\), since \(2^2+1^2=5\). Exam tip: the hypotenuse lies opposite the right angle.

A square root spiral is formed using successive right triangles. The hypotenuse is found by Pythagoras’ theorem, \(c^2=a^2+b^2\), and represents the next square root. Exam tip: identify the hypotenuse first.

A square root spiral is a geometric construction based on successive right triangles. At each stage, a segment of length 1 is drawn perpendicular to the previous hypotenuse. If the previous hypotenuse is √n, the Pythagorean theorem gives the next one as √((√n)² + 1²) = √(n + 1). Repeating this process produces lengths representing successive square roots, such as √2, √3, √4, and beyond. Therefore option A states the central idea accurately. The construction is not a circle-only activity, so B is incorrect. It may help illustrate number concepts but does not merely calculate decimals, making C incorrect. It produces positive lengths, not only negative numbers, so D is also incorrect.

To show the length on the number line, the compass point is placed at the origin and an arc is drawn. This gives the correct square root point.

A square root spiral begins with a line segment of length 1 unit. A perpendicular segment of 1 unit is then drawn at one end to form the first right triangle, whose hypotenuse is \(\sqrt{2}\). Thus, \(\sqrt{2}\) is obtained after the first construction and is not the starting segment. Exam tip: Remember that the spiral always starts with a 1-unit base.

The governing concept is the Pythagorean theorem. For a right triangle with perpendicular sides a and b and hypotenuse h, the relationship is h²=a²+b². Here both perpendicular sides are 1 unit, so h²=1²+1²=1+1=2. Since a length is positive, h=√2 units. Therefore option B is correct. √1 represents the length of one individual unit side, not the hypotenuse formed by the two perpendicular sides. √3 and √4 are later square-root values that could occur after additional steps in the spiral, but they do not describe this particular triangle. Thus the first construction from two unit perpendicular lengths gives √2.

In a square root (Theodorus) spiral, each new right triangle is constructed on the previous hypotenuse, with a new perpendicular side of \(1\) unit. By the Pythagorean theorem, the successive hypotenuses become \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Thus, \(\sqrt{2}\) and \(\sqrt{3}\) are hypotenuses at early stages, not the new perpendicular side. Exam tip: remember that the side kept fixed in every new triangle is \(1\) unit.